LCT经验总结

妈妈，我终于会写LCT了！
LCT太神奇了，它和Splay能完美结合。在$O(nlogn)$的时间内解决动态树问题。

BZOJ2049 [Sdoi2008]Cave 洞穴勘测

丢板子，懒得解释。用维护一下联通块就行了，因为数据水，并查集也可以水过。

AC代码：

#include <iostream>
#include <cstdio>
#define ls ch[0]
#define rs ch[1]
#define lson s[x].ls
#define rson s[x].rs
#define father s[x].fa
const int MAXN = 10000;
using namespace std;

inline int read() {
    int x = 0, w = 1; char c = ' ';
    while (c < '0' || c > '9') { c = getchar(); if (c == '-') w = -1; }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x * w;
}

int n = read(), m = read();

class LCT {
    struct Node {
        int data, ch[2], fa;
        bool rev;
    } s[MAXN + 5];
    public :
        LCT() { top = 0; }//, for (int i = 1; i <= n; ++i) s[i].data = read(); }
        void PushDown(int x) {  
            if (s[x].rev) {
                s[lson].rev ^= 1, s[rson].rev ^= 1, swap(lson, rson);
                s[x].rev ^= 1;
            }
        }
        bool IsRoot(int x);
        void rotate(int x); void Splay(int x);
        void access(int x); void MakeRoot(int x); 
        void link(int x, int y); void cut(int x, int y);
        int find(int x);
    private :
        int top, ST[MAXN + 5];
} T;

bool LCT::IsRoot(int x) {
    return s[father].ls != x && s[father].rs != x; 
}

void LCT::rotate(int x) {
    int y = father, z = s[y].fa, l, r;
    if (s[y].ls == x) l = 0; else l = 1;
    r = l ^ 1;
    if (!IsRoot(y))
        if (s[z].ls == y) s[z].ls = x; else s[z].rs = x;
    father = z, s[y].fa = x, s[ s[x].ch[r] ].fa = y, s[y].ch[l] = s[x].ch[r], s[x].ch[r] = y; 
}

void LCT::Splay(int x) {
    top = 0, ST[++top] = x;
    for (int i = x; !IsRoot(i); i = s[i].fa) ST[++top] = s[i].fa;
    for (int i = top; i; --i) PushDown(ST[i]);
    while (!IsRoot(x)) {
        int y = father, z = s[y].fa;
        if (!IsRoot(y)) 
            if (s[y].ls == x ^ s[z].ls == y) rotate(x); else rotate(y); 
        rotate(x);
    }
}

void LCT::access(int x) {
    for (int last = 0; x; last = x, x = father) 
        Splay(x), rson = last;
}

void LCT::MakeRoot(int x) {
    access(x), Splay(x), s[x].rev ^= 1;
}

void LCT::link(int x, int y) {
    MakeRoot(x), father = y;
}

void LCT::cut(int x, int y) {
    MakeRoot(x), access(y), Splay(y); 
    if (s[y].ls == x) s[y].ls = father = 0;
}

int LCT::find(int x) {
    access(x), Splay(x);
    while (lson) x = lson;
    return x;
}

char c[20];

int main() {
    for (int i = 1; i <= m; ++i) {
        scanf ("%s", c); int u = read(), v = read();
        switch(c[0]) {
            case 'Q' : if (T.find(u) == T.find(v)) puts("Yes"); else puts("No"); break;
            case 'D' : T.cut(u, v); break;
            case 'C' : T.link(u, v); break;
        }
    }   
} 

BZOJ3282 Tree

顺便吐槽一下，BZOJ太多题都叫Tree了。又是动态树板子题。是不是动态树都是板子题啊。用一个要上传标记的Splay维护即可。

AC代码：

#include <iostream>
#include <cstdio>
#define ls ch[0]
#define rs ch[1]
#define lson s[x].ls
#define rson s[x].rs
#define father s[x].fa
const int MAXN = 3e5;
using namespace std;
  
inline int read() {
    int x = 0, w = 1; char c = ' ';
    while (c < '0' || c > '9') { c = getchar(); if (c == '-') w = -1; }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x * w;
}
  
int n = read(), m = read();
  
class LCT {
    struct Node {
        int data, ch[2], fa, val;
        bool rev;
    } s[MAXN + 5];
    public :
        LCT() { for (int i = 1; i <= n; ++i) s[i].data = read(); }
        void PushUp(int x) {
            s[x].val = s[lson].val ^ s[rson].val ^ s[x].data;
        }
        void PushDown(int x) {
            if (s[x].rev) {
                s[x].rev ^= 1, s[lson].rev ^= 1, s[rson].rev ^= 1;
                swap (lson, rson);
            }
        }
        bool IsRoot(int x);
        void rotate(int x); void Splay(int x);
        void MakeRoot(int x); void access(int x); 
        void link(int x, int y); void cut(int x, int y);
        int find(int x); void update(int x, int cnt); int query(int x, int y);
    private :
        int top, ST[MAXN + 5];
} T;
  
bool LCT::IsRoot(int x) {
    return s[father].ls != x && s[father].rs != x;
}
  
void LCT::rotate(int x) {
    int y = father, z = s[y].fa, l, r;
    if (s[y].ls == x) l = 0; else l = 1; 
    r = l ^ 1;
    if (!IsRoot(y))
        if (s[z].ls == y) s[z].ls = x; else s[z].rs = x;
    father = z, s[y].fa = x, s[ s[x].ch[r] ].fa = y, s[y].ch[l] = s[x].ch[r], s[x].ch[r] = y;
    PushUp(y), PushUp(x);
}
  
void LCT::Splay(int x) {
    top = 0, ST[++top] = x; 
    for (int i = x; !IsRoot(i); i = s[i].fa) ST[++top] = s[i].fa; 
    for (int i = top; i; --i) PushDown(ST[i]);
    while (!IsRoot(x)) {
        int y = father, z = s[y].fa;
        if (!IsRoot(y))
            if (s[y].ls == x ^ s[z].ls == y) rotate(x); else rotate(y);
        rotate(x);
    }
}
 
void LCT::access(int x) {
    for (int last = 0; x; last = x, x = father)
        Splay(x), rson = last, PushUp(x);
}
  
void LCT::MakeRoot(int x) {
    access(x), Splay(x), s[x].rev ^= 1;
} 
 
void LCT::link(int x, int y) {
    MakeRoot(x), father = y;
}
 
void LCT::cut(int x, int y) {
    MakeRoot(x), access(y), Splay(y);
    if (s[y].ls == x) s[y].ls = father = 0;
}
 
int LCT::find(int x) {
    access(x), Splay(x);
    while (lson) x = lson;
    return x;
}
 
void LCT::update(int x, int cnt) {
    access(x), Splay(x), s[x].data = cnt, PushUp(x);
}
 
int LCT::query(int x, int y) {
    MakeRoot(x), access(y), Splay(y);
    return s[y].val;
} 
 
int main( ) {
    for (int i = 1; i <= m; ++i) {
        int opt = read(), u = read(), v = read();
        switch (opt) {
            case 0 : printf("%d\n", T.query(u, v) ); break;
            case 1 : if (T.find(u) != T.find(v)) T.link(u, v); break;
            case 2 : if (T.find(u) == T.find(v)) T.cut(u, v); break;
            case 3 : T.update(u, v);
        }
    }
    return 0;
}

经验总结

拒绝压行，从我做起（大雾）。

有时间再填坑吧